Simplify the following expression: $y = \dfrac{9x^2+10x+1}{9x + 1}$
Answer: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(9)}{(1)} &=& 9 \\ {a} + {b} &=& &=& {10} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $9$ and add them together. The factors that add up to ${10}$ will be your ${a}$ and ${b}$ When ${a}$ is ${1}$ and ${b}$ is ${9}$ $ \begin{eqnarray} {ab} &=& ({1})({9}) &=& 9 \\ {a} + {b} &=& {1} + {9} &=& 10 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({9}x^2 +{1}x) + ({9}x +{1}) $ Factor out the common factors: $ x(9x + 1) + 1(9x + 1)$ Now factor out $(9x + 1)$ $ (9x + 1)(x + 1)$ The original expression can therefore be written: $ \dfrac{(9x + 1)(x + 1)}{9x + 1}$ We are dividing by $9x + 1$ , so $9x + 1 \neq 0$ Therefore, $x \neq -\frac{1}{9}$ This leaves us with $x + 1; x \neq -\frac{1}{9}$.